Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
   public static void main(String[] args)  {
        try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
            br.readLine();
            String[] split = br.readLine().split(" ");
            int n = split.length;
            double[] prob = new double[n];
            for (int i = 0; i < n; i++) {
               prob[i] = Double.parseDouble(split[i]); 
            }
            double[] dp = new double[n + 1];
            dp[0] = 1.0;
            for (int i = 1; i <= n; i++) {
                for (int j = Math.min(i + 1, n); j >= 0; j--) {
                    dp[j] = (j == 0 ? 0 : dp[j - 1])*prob[i - 1] +  dp[j]*(1 - prob[i - 1]);
                }
            }
            double result = 0.0;
            for (int i = (n + 1) / 2 ; i <= n; i++) {
                result += dp[i];
            }
            System.out.println(result);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}